Question

When 25.0 mL of 0.500 M H2SO4 is added to 25.00 mK of 1.00 M KOH in a coffee-cup calorimeter at 23.50◦ C, the temperature rises to 30.17◦ C. Calculate DHrxn for this reaction. (Assume that the density of water of and specific heat capacity of the solution are the same as for pure water). (Answer: –55.8 kJ/mol KOH, –112 kJ/mol H2SO4)

Answer 1

`\Delta H_rxn=-55.8(kJ)/(mol KOH)`

`\Delta H_rxn=-111.6(kJ)/(mol H2SO4)`

Step-by-step explanation:

The reaction:
`H_2SO_4 (aq) + 2 KOH (aq) \longrightarrow K_2SO_4 (aq) + H_2O (l)`

Total volume: V=50 mL

The heat generated:

`Q=Cp_(water)*\Delta T*V*\delta_(water)`

`Q=4.184 (J)/(g*C)*(30.17 C -23.5 C)*50 mL*(1 g)/(mL)}`

`Q=1395.4 J`

Moles of KOH:
`n_K=(1 mol)/(1000 mL)*25 mL=0.025 mol`

Moles of H2SO4:
`n_S=(0.5 mol)/(1000 mL)*25 mL=0.0125 mol`

Heat of reaction:

`\Delta H_(rxn)=(-1395.4 J)/(0.025 mol KOH)*(1 kJ)/(1000 J)`

`\Delta H_(rxn)=-55.8(kJ)/(mol KOH)`

`\Delta H_(rxn)=(-1395.4 J)/(0.0125 mol H2SO4)*(1 kJ)/(1000 J)`

`\Delta H_(rxn)=-111.6(kJ)/(mol H2SO4)`

Note: the negative sign is beacuse it is an exothermical reaction.